# What's the relation between output voltage and time to boil water given the same kettle?

An electric kettle rated 220V, 2000W needed 10 minutes to boil water when it is half filled with water in Singapore where the output voltage is 220V. Estimate the amount of time needed to do the same task if the kettle was brought to the USA where the output voltage is 110V.

Options:

1. 5 minutes
2. 10 minutes
3. 20 minutes
4. 40 minutes


To be able to answer this you have to know how the output voltage is related to the power produced by the kettle. Can someone explain this relationship?

I don't quite understand what it is meant by " An electric kettle rated 220V, 2000W". I get that 2000W means the kettle uses 2000J/s and V is the work done by kettle per unit charge. But what is the difference between the 2 (power and charge of the kettle)? And how does the V of kettle differ from output voltage?

• I think the main thing you're asking here, namely how to derive the 40-minute answer, is off topic according to our homework policy. But your followup (P.S.) question might be on topic, I'm not sure. That could be more of a conceptual question. Sep 12 '14 at 23:48
• @DavidZ I am aware of the HW policy. I disagree with your application & deletion of the post in this case. (1) My answer provides conceptual insights (2) the OP already has the answer key, and got it wrong. They're clearly missing something conceptually hence the post in the first place. Sep 13 '14 at 3:43
• @DavidZ Those points do speak to the (in)appropriateness of deleting the post, actually. The whole motivation of the clause in the HW policy of providing complete answers, IMO, is to prevent cheating. Since that isn't an issue here, that point is moot. Sep 13 '14 at 12:32
• @user3814483 I maintain that they don't, but again, let's move this to Physics Meta or Physics Chat if you want to discuss further. This isn't the place. I can also try to explain in another venue how deleting your answer fits into our larger goal of preventing people from cheating on their homework or exams here. Sep 13 '14 at 23:28
• Happytreat, I've edited your question in an attempt to get it reopened. If you don't like my edit you can undo my changes from the revisions page. Sep 14 '14 at 11:13

First, this is a poor question (the question that was asked of you, presumably). That's because it requires you to make an assumption about how the kettle works.

For instance, if the kettle's circuitry strives to maintain a constant power output (2000 W), then it will draw more current and maintain its 2000 W output when operated at 110V, which would mean no change in the time.

Since 10 minutes is not the correct answer (and if I were grading this test and someone said 10 minutes I'd say that's acceptable, because I let you assume how it operates). One assumption is to say that the heating element inside the kettle is fixed ($R$ = const) such that the power dissipated goes like $V^2 / R$. With half the voltage, you'd have 1/4 the power, and loosely speaking, it would take 4 times as long, so 40 minutes.

Your teacher should have added, "assume the kettle is nothing more than a resistor that has a voltage applied across it."

The key think you need to know is that if you apply a voltage $V$ and a current $I$ flows then the power dissipated is:

$$P = VI \tag{1}$$

You know the power at 220V so you can calculate the current, and if we assume the kettle behaves as a simple resistor then it will obey Ohm's law so you can calculate the resistance.

$$R = \frac{V}{I} \tag{2}$$

Now you can work out how the current changes when you reduce the voltage from 220V to 110V, and then use equation (1) to calculate the power. If you work out the factor by which the power has decreased you can divide by this factor to get the new time to boil the water.

You can take a short cut by rearranging equation (2) to read:

$$I = \frac{V}{R}$$

If you substitute this expression for $I$ in equation (1) you get:

$$P = VI = V\frac{V}{R} = \frac{V^2}{R}$$

Assuming the resistance $R$ is a constant we find that power is proportional to voltage squared. So if you reduce the voltage by a factor $f$ you reduce the power by a factor $f^2$.

We should probably note that in eal life the resistance $R$ almost certainly isn't the same at 220V as at 110V because resistance generally depends on temperature and the heating element won't be as hot at 110V.